Chapter 11, Problem 41P

(a)

To determine

The radius of the orbit of the hydrogen atom in the first excited state.

Answer to Problem 41P

The radius of the orbit of the hydrogen atom in the first excited state is $\underset{\_}{0.212\text{nm}}$.

Explanation of Solution

Write the expression for the radius of the orbit in the hydrogen atom relating the Bohr radius.

    $r_n=a_0{n}^2$        (I)

Here, $r_n$ is the radius of the orbit of the $n^{\text{th}}$ state, $a_0$ is the Bohr radius, $n$ is the state.

Conclusion:

Substitute $0.0529\text{nm}$ for $a_0$, $2$ for $n$ in equation (I) to find $r_2$.

    $\begin{array}{c}{r}_2=\left(0.0529\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right){\left(2\right)}^2\\ =0.212\text{nm}\end{array}$

Therefore, the radius of the orbit of the hydrogen atom in the first excited state is $\underset{\_}{0.212\text{nm}}$.

(b)

To determine

The linear momentum of the electron in the hydrogen atom.

Answer to Problem 41P

The linear momentum of the electron in the hydrogen atom is $\underset{\_}{9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the columbic force of attraction.

    $F_C=\frac{k_e{e}^2}{r^2}$        (II)

Here, $F_C$ is the columbic force, $k_e$ is the columbic constant, $e$ is the electronic charge.

Write the expression for the centripetal force of the electrons revolving in the orbit.

    $F=\frac{m_e{v}_2^2}{r_2}$        (III)

Here, $F$ is the centripetal force, $m_e$ is the mass of the electron, $v_2$ is the velocity in the $n=2$ state.

Equate equation (II) and (III) to solve for ${\left(m_e{v}_2\right)}^2$.

    ${\left(m_e{v}_2\right)}^2=\sqrt{\frac{m{k}_e{e}^2}{r}}$        (IV)

Write the expression for the linear momentum.

    $P_2=m_e{v}_2$        (V)

Here, $P_2$ is the momentum in the $n=2$ state.

Use equation (IV) to solve for $P_2$.

    $P_2=\sqrt{\frac{m_e{k}_e{e}^2}{r_2}}$        (VI)

Conclusion:

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $1.60\times {10}^{-19}\text{C}$ for $e$, $0.212\text{nm}$ for $r_2$ in equation (VI) to find $P_2$.

    $\begin{array}{c}{P}_2=\sqrt{\frac{\left(9.11\times {10}^{-31}\text{kg}\right)\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{\left(0.212\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}}\\ =9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Therefore, The linear momentum of the electron in the hydrogen atom is $\underset{\_}{9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}}$.

(c)

To determine

The angular momentum of the electron.

Answer to Problem 41P

The angular momentum of the electron is $\underset{\_}{2.11\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}}$.

Explanation of Solution

Write the expression for the angular momentum.

    $L_2=P_2{r}_2$        (VII)

Here, $L_2$ is the angular momentum.

Conclusion:

Substitute $9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}$ for $P_2$, $0.212\text{nm}$ for $r_2$ in equation (VII) to find $L_2$.

    $\begin{array}{c}{L}_2=\left(9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}\right)\left(0.212\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =2.11\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Therefore, the angular momentum of the electron is $\underset{\_}{2.11\times {10}^{-34}\text{kg}\cdot {\text{m}}^2/\text{s}}$.

(d)

To determine

The kinetic energy of the electron.

Answer to Problem 41P

The kinetic energy of the electron is $\underset{\_}{5.43\times {10}^{-19}\text{J}}$.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

    $K_2=\frac{1}{2}{m}_e{v}_2^2$        (VIII)

Use equation (IV) to solve for $K_2$.

    $K_2=\frac{P_2^2}{2m_e}$        (IX)

Conclusion:

Substitute $9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}$ for $P_2$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$ in equation (IX) to find $K_2$.

    $\begin{array}{c}{K}_2=\frac{{\left(9.95\times {10}^{-25}\text{kg}\cdot \text{m}/\text{s}\right)}^2}{2\left(9.11\times {10}^{-31}\text{kg}\right)}\\ =5.43\times {10}^{-19}\text{J}\\ K_2\left(\text{in eV}\right)=\frac{5.43\times {10}^{-19}\text{J}}{1.60\times {10}^{-19}\text{C}}\\ =3.40\text{eV}\end{array}$

Therefore, the kinetic energy of the electron is $\underset{\_}{5.43\times {10}^{-19}\text{J}}$.

(e)

To determine

The potential energy of the electron.

Answer to Problem 41P

The potential energy of the electron is $\underset{\_}{-6.80\text{eV}}$.

Explanation of Solution

Write the expression for the potential energy.

    $U_2=-\frac{k_e{e}^2}{r_2}$        (X)

Here, $U_2$ is the potential energy of the electron.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $1.60\times {10}^{-19}\text{C}$ for $e$, $0.212\text{nm}$ for $r_2$ in equation (X) to find $U_2$.

    $\begin{array}{c}{U}_2=-\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{\left(0.212\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =-1.09\times {10}^{-18}\text{J}\\ U_2\left(\text{in eV}\right)=\frac{-1.09\times {10}^{-18}\text{J}}{1.60\times {10}^{-19}\text{C}}\\ =-6.80\text{eV}\end{array}$

Therefore, the potential energy of the electron is $\underset{\_}{-6.80\text{eV}}$.

(f)

To determine

The total energy of the system.

Answer to Problem 41P

The total energy of the system is $\underset{\_}{-3.40\text{eV}}$.

Explanation of Solution

Write the expression for the total energy of the system.

    $E_2=K_2+U_2$        (XI)

Here, $E_2$ is the total energy of the system.

Conclusion:

Substitute $3.40\text{eV}$ for $K_2$, $-6.80\text{eV}$ for $U_2$ in equation (XI) to find $E_2$.

    $\begin{array}{c}{E}_2=3.40\text{eV}-6.80\text{eV}\\ \text{=-3}\text{.40}\text{eV}\end{array}$

Therefore, the total energy of the system is $\underset{\_}{-3.40\text{eV}}$.